Circles

3. A small radio transmitter broadcasts in a 50 mile radius. If you drive along a straight line from a city 60 miles north of the transmitter to a second city 70 miles east of the transmitter, during how much of the drive will you pick up a signal from the transmitter?

I am completely lost on how I'm supposed to go about solving this problem. The solution was provided with brief steps but I still couldn't wrap my head around it. Thank you!

Re: Circles

Originally Posted by xSummer

3. A small radio transmitter broadcasts in a 50 mile radius. If you drive along a straight line from a city 60 miles north of the transmitter to a second city 70 miles east of the transmitter, during how much of the drive will you pick up a signal from the transmitter?

I am completely lost on how I'm supposed to go about solving this problem. The solution was provided with brief steps but I still couldn't wrap my head around it. Thank you!

-Summer

You can see from the image that you intersect the circle of radio range in two spots.

Let the tower be at (0,0). City 1 is then at (0,60) and city 2 at (70,0).

Consider the line connecting the two cities.

The slope of this line is $m=\dfrac{(0-70)}{(60-0)}=- \dfrac 7 6$

The equation of the line is thus

$(y-60) = -\dfrac 7 6 x \Rightarrow y = -\dfrac 7 6 x$

In the quadrant where the trip takes place the equation for the circle of radio range is

$y = \sqrt{50^2 - x^2}=\sqrt{2500 - x^2}$

Now just solve for the two intersections

$\sqrt{2500 - x^2}=-\dfrac 7 6 x$

can you do this? You'l get two solutions corresponding to the two point on the circle. Find the distance between them. Find the distance between the cities. Take the ratio of these two distances and you're done.

Re: Circles

You can see from the image that you intersect the circle of radio range in two spots.

Let the tower be at (0,0). City 1 is then at (0,60) and city 2 at (70,0).

Consider the line connecting the two cities.

The slope of this line is $m=\dfrac{(0-70)}{(60-0)}=- \dfrac 7 6$

The equation of the line is thus

$(y-60) = -\dfrac 7 6 x \Rightarrow y = -\dfrac 7 6 x$

Typo here. What's meant is

$y=-\dfrac 7 6 x + 60$

In the quadrant where the trip takes place the equation for the circle of radio range is

$y = \sqrt{50^2 - x^2}=\sqrt{2500 - x^2}$

Now just solve for the two intersections

$\sqrt{2500 - x^2}=-\dfrac 7 6 x$

and again what's meant is

$\sqrt{2500 - x^2}=-\dfrac 7 6 x + 60$

can you do this? You'l get two solutions corresponding to the two point on the circle. Find the distance between them. Find the distance between the cities. Take the ratio of these two distances and you're done.

Re: Circles

Hello, xSummer!

Did you make a sketch?

3. A small radio transmitter broadcasts in a 50 mile radius.
If you drive along a straight line from a city 60 miles north of the transmitter
to a second city 70 miles east of the transmitter, during how much of the drive
will you pick up a signal from the transmitter?

On your graph, draw a circle centered at the Origin with radius 50.
Draw a line from (0,60) to (70,0).
Note where the line intersects the circle, points A and B.

Chord AB represents the locations where you can receive the broadcast.